Prove that the length of the common chord of the two circles #x^2+y^2=a^2# and #(x-c)^2+y^2=b^2# is #1/c(sqrt((a+b+c)(a-b+c)(a+b-c)(-a+b+c)))#, where #a,b,c>0#?

1 Answer
P dilip_k
Nov 3, 2017

The equations of the circles given are

#C_1->x^2+y^2=c^2.....(1)#

#C_2->(x-c)^2+y^2=b^2.....(2)#

Subtracting (2) from (1) we get the equation of common chord as

#2cx-c^2=a^2-b^2#

#=>x=1/(2c)(a^2-b^2+c^2)#

This means the common chord is a straight line parallel to y-axis.

So the end points of the common chord will have same x-coordinates i.e.#1/(2c)(a^2-b^2+c^2)# and the y-coordinates of the end points can be obtained by putting # x=1/(2c)(a^2-b^2+c^2)# in equation (1) or (2).

So the y-coordinates of the end points of the common chord will be

#pmsqrt(a^2-(1/(2c)(a^2-b^2+c^2))^2#

Hence the length of the chord will be
#=2sqrt(a^2-(1/(2c)(a^2-b^2+c^2))^2#

#=2sqrt((4a^2c^2-(a^2-b^2+c^2)^2)/(4c^2))#

#=2/(2c)sqrt(((2ac)^2-(a^2-b^2+c^2)^2))#

#=1/csqrt((2ac+a^2-b^2+c^2)(2ac-a^2+b^2-c^2))#

#=1/csqrt(((a+c)^2-b^2)(b^2-(a-c)^2))#

#=1/csqrt((a+b+c)(a-b+c)(a+b-c)(-a+b+c))#

Proved

Related questions
Impact of this question
3746 views around the world
You can reuse this answer
Creative Commons License